JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2017
- Medium
- Numerical
A capacitance of \(2\ μF\) is required in an electrical circuit across a potential difference of \(1.0\ kV\). A large number of \(1\ μF\) capacitors are available which can withstand a potential difference of not more than \(300\ V\). The minimum number of capacitors required to achieve this is
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Answer & Solution
Correct Answer
(D) \(32\)
Step-by-step Solution
Detailed explanation
To get a capacitance of \(2\, \mu \mathrm{F}\) arrangement of capacitors of capacitance \(1\, \mu \mathrm{F}\) as shown in figure \(8\) capacitors of \(1\, \mu \mathrm{F}\) in parallel with four such branches in series i.e., \(32\) such capacitors are required.
\(\frac{1}{C_{e q}}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8} \quad \therefore C_{e q}=2\, \mu F\)
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